ECON 381 SC 2012 ASSIGNMENT 1 ANSWERS Answer all the ...

ECON 381 SC 2012 ASSIGNMENT 1ANSWERSJOHN HILLASUNIVERSITY OF AUCKLANDAnswer all the following problems. Each problem is worth the same number ofmarks (20). For questions with multiple parts each part is worth the same numberof marks. Some parts are however significantly harder and longer than others.Problem 1. Use a truth table to find whether the following statements are equivalent.Explain precisely why the truth table implies the result that you state.((p ⇒ r)∨(q ⇒ p)) and ((p∧q) ⇒ r).p q r ((p ⇒ r)∨(q ⇒ p)) ((p∧q) ⇒ r)T T T T TT T F T FT F T T TT F F T TF T T T TF T F T TF F T T TF F F T TWe see on the second line of the truth table that when p is true,q is true and r is false the first statement is true while the secondstatement is false. So they are not logically equivalent. In fact thefirst statement is a tautology.Problem 2. Let X = {a,b,c,d,e}. In the following questions use this set X.(i) Write out explicitly what the set X×X is. Explain what we mean by a binaryrelation on X and define this explicitly and formally.X ×X = {(a,a),(a,b),(a,c),(a,d),(a,e),(b,a),(b,b),(b,c),(b,d),(b,e),(c,a),(c,b),(c,c),(c,d),(c,e),(d,a),(d,b),(d,c),(d,d),(d,e),(e,a),(e,b),(e,c),(e,d),(e,e)}A binary relation on X expresses some relation among themembers of X. We write xRy or (x,y) ∈ R to mean thatx has the relation R to y. Formally a binary relation is anysubset of X ×X, including the empty set and all of X ×X.(ii) Give an example of a binary relation on X that is complete and one that isnot complete.Date: Second Semester, 2011.1

2 JOHN HILLAS UNIVERSITY OF AUCKLANDThere are many correct answers to this question. The essentialfeature of a correct answer is that for each pair of pointsx and y either (x,y) or y,x should be in the set R. For exampleeither (a,b) or (b,a) should be in the set. Since in ourdefinition of completeness we do not exclude the possibilitythat x = y we also require that for each possible value of xthe pair (x,x) is in the set R. A relation is not complete if forsome pair of points x and y neither (x,y) nor (y,x) is in theset R. Perhaps the simplest, though not the most interestinganswers are that R = X × X is complete and R = ∅ is notcomplete. But there are many other correct answers.(iii) Give an example of a binary relation that is transitive but not negativelytransitive.Again, there are many correct answers. One particularlymeaningful one that we discussed in class was the strict preferencerelation for a consumer with threshold effect. Supposeor consumer likes earlier letters in the alphabet better thanlater letters, but in order to actually prefer one letter overanotherit must be at least two places earlier. Thus he prefersa to c but is indifferent between a and b. Such preferenceswould be described by the following relation:P = {(a,c),(a,d),(a,e),(b,d),(b,e),(c,e)}.This relation is transitive. The only possibility for x, y, andz all different with xPy and yPz is x = a, y = c and z = e,but in this case the condition for transitivity is met since wehave aPe, that is (a,e) is in P. However the relation is notnegatively transitive since we have aPc but have neither aPbnor bPc.Another relation that is transitive but not negatively transitiveis given by any nontrivial equivalence. In our example“equals” is a natural example.I = {(a,a),(b,b),(c,c),(d,d),(e,e)}.Again, this is transitive. The only possibility for x, y, andz with xIy and yIz has x = y = z, but then we also havexIz. It is not negatively transitive since we have aIa buthave neither aIb nor bIa.You can make slightly less trivial examples by having theconsumer indifferent between some pairs of letters. ConsiderI = {(a,a),(a,b),(b,a),(b,b),(c,c),(c,d),(d,c),(d,d),(e,e)}.This relation too is transitive, but not negatively transitive.Proof for this case is left to you.(iv) Give an example of a binary relation that is negatively transitive but nottransitive.The notion of negative transitivity of a relation R is so calledbecause it corresponds to the relation defined by notR beingtransitive. Thus each of the examples in the previous answergenerates an answer to this question. Simply take the complementof the relation in X ×X. So for P this would give

ECON 381 SC 2012 ASSIGNMENT 1 ANSWERS 3the relationnotP = X ×X\P = {(a,a),(a,b),(b,a),(b,b),(b,c),(c,a),(c,b),(c,c),(c,d),(d,a),(d,b),(d,c),(d,d),(d,e),(e,a),(e,b),(e,c),(e,d),(e,e)}This relation is not transitive since we have (a,b) and (b,c)in the relation but not (a,c). It’s a bit more tedious to seethat the relation is negatively transitive. For notP to benegatively transitive we require that if xnotPy then eitherxnotPz or znotPy. We see that there are quite a few pairsin the relation notP. It would be very tedious to check eachcase that xnotPy. But we know that the condition will besatisfied as long as either xnotPz or znotPy and it’s a loteasier to check the cases in which this does not happen. Forwhat values of x, y, and z do we have not ((xnotPz) or(znotPy)) that is (not(xnotPz)) and (not(znotPy))? Fromthe definition of notP this is when xPz and zPy. As we sawin the answer to the previous question the only possibility forthis is x = a, z = c, and y = e. But then (x,y) = (a,e) is inP and so is not in notP, and so for this combination of x, y,and z we do not require that either xnotPz or znotPy. Thuswe have the condition “either xnotPz or znotPy” wheneverrequired and so the relation notP is negatively transitive.Similarlywecouldgenerateexamplesfromanyoftheotheranswers to the previous question.(v) What is an equivalence? Give an example of an equivalence on X.An equivalence is a binary relation that is symmetric andtransitive. We gave a couple of examples in the answer toquestion (ii), namelyandI = {(a,a),(b,b),(c,c),(d,d),(e,e)}I = {(a,a),(a,b),(b,a),(b,b),(c,c),(c,d),(d,c),(d,d),(e,e)}.Since we have already seen that these relations are transitivewe need only to check that they are symmetric, that isthat when (x,y) is in the relation then so is (y,x). This iscompletely trivial in the first case—the only pairs in the relationare of the form (x,x). In the second case the additionalpairs are (a,b) and (b,a) and (c,d) and (d,c). so we see thatwhenever (x,y) is there then so is (y,x), as required.Problem 3. Let X = {x,y,z} and Y = {a,b,c,d,e}(i) Give an example of a function f : X → Y that is one-to-one, that is, that isan injection.The requirement is that different points in X go to differentpoints in Y. There are many such functions. One is given byf(x) = a, f(y) = b, and f(z) = c.(ii) Is the function that you have given onto, that is, a surjection?

4 JOHN HILLAS UNIVERSITY OF AUCKLANDNo. ThereisnoelementofX thatmapstod. Weseethatthismust be the case whatever function we choose there are onlythree points in X and each of them maps to only one pointin Y. Since there are 5 points in Y at least 2 of them mustnot be images of any point in X. (Given that the function isassumed to be one-to-one exactly two of them are not imagesof any point in X. In my answer d and e are not images ofany point in X.(iii) What is the range of the function you have given?The range of f, that is f(X) is {a,b,c}(iv) If we let the range of f be denoted f(X) consider the function ˜f : X → f(X)?Is this function one-to-one or onto (or both)?It’s one-to-one since the function f is one-to-one. It’s ontosince all points that are not images of some point in X havebeen left out of the co-domain.(v) Does ˜f have an inverse? If so find that inverse, if not explain why the inversedoes not exist.Yes. Being one-to-one and onto is a necessary and sufficientcondition for a function to have an inverse. ˜f−1 : f(X) → Xis given by ˜f −1 (a) = x, ˜f−1 (b) = y, and ˜f −1 (c) = z.Problem 4. Consider the following “metric” (“metric” is in quotes since, until youdo part (i) of the problem, you don’t know that this is actually a metric) on theset of all n−tuples of real numbers:{0 if x = y,d D (x,y) =1 otherwise.This metric is called the discrete metric.(i) Prove that for any set X the discrete metric d D is a metric.A function d : X ×X → R is a metric if for all x, y, and z inX1. d(x,y) ≥ 0 and d(x,y) = 0 if and only if x = y,2. d(x,y) = d(y,x),3. d(x,y) ≤ d(x,z)+d(z,y).If x = y then d D (x,y) = 0 = d D (y,x) and if x ≠ y thend D (x,y) = 1 = d D (y,x). This proves both properties 1 and2.If x = y then d D (x,y) = 0 ≤ d D (x,z)+d D (z,y) by Property1,whichwe’vealreadyproved.Ifx ≠ y thend D (x,y) = 1but we must have either x ≠ z or y ≠ z (since if x = z andy = z then x = y). But then d D (x,z) + d D (z,y) is either1 + 0 = 1 or 0 + 1 = 1 or 1 + 1 = 2, and, in any cased D (x,y) = 1 ≤ d D (x,z) + d D (z,y). Hence Property 3 issatisfied.(ii) Let X be any set and consider the metric space (X,d D ). What are the opensubsets of X with this metric? You should prove your answer not just statewhat the set is.What are the open balls with metric d D ? If the radius r isgreaterthan 1 then for any x ∈ X is distance less than r fromx so every point in X is in the ball. If the radius is less thanor equal to 1 then onlyx is distance less than r from x. That

ECON 381 SC 2012 ASSIGNMENT 1 ANSWERS 5isB r (x) ={{x} if r ≤ 1X if r > 1 .Now let S ⊂ X and let x ∈ S. Then there is some r > 0,namely r = 1 (or any positive value less than 1), such thatB r (x) = {x} ⊂ S. Thus S is an open set.We have shown that any set S ⊂ X is open. The open setsare all the subsets of X.Notice that we didn’t make any assumptions about the setX. We can define the discrete metric on any set and withthat metric all the subsets of our set are open. This usuallyisn’t a particularly useful structure to put on the set.Problem 5. Let (X,d) be a metric space.Another approach to putting structure of a set X is to directly specify the opensets. Such a space is called a topological space. If (X,τ) is a topological space thenX is a set and τ is a collection of subsets of X such that1. X ∈ τ2. ∅ ∈ τ3. If A and B are in τ then A∩B is in τ. That is, the intersection of two elementsof τ is in τ. (This also implies that the intersection of any finite number ofelements of τ is in τ.)4. If A n is in τ for every n ∈ N for some set N that might be finite or infinite then∪ n∈N A n is in τ. That is, the union of any number of elements of τ is in τ.Now let X = {a,b,c,d,e}. For the questions below use this particular set X.(i) Show that for any metric d the set of open subsets of X are the same. Findwhat this collection of open sets is.In the previous problem we considered a general set butlooked at a particular metric for each possible set. In thisproblem we look at a particular set X but ask about all possiblemetrics. The essential feature of the set we consider isthat it contains only a finite number of elements—in our casefive. We essentially find the same result. All sets are open.And the reasoningis quite similar. Let us consider some metricd on X. For any x and y in X with x ≠ y d(x,y) > 0.Consider s = min x,y∈X|x≠y d(x,y), that is the smallest distancebetween elements of X other than the distance fromeach point to itself. This is the smallest of a finite umberof strictly positive numbers and so s > 0 and if x ≠ y thend(x,y) ≥ s. Thus for any x ∈ X = {a,b,c,d,e} we haveB s (x) = {x}. Now, as in the previous problem, let S ⊂ Xand let x ∈ S. Then there is some r > 0, namely r = s (orany positive value less than s), such that B r (x) = {x} ⊂ S.Thus S is an open set. Again the set of open sets is the collectionof all subsets of X. (Of course in this case we knowwhat X is so we know what the set of all subsets is. Thereare 32 of them, including the empty set.(ii) Show that if we let τ = {X,∅} that this collection of “open sets” satisfies thefour properties given above.This is quite an easy problem. The only difficulty is beingpedantic enough to find something to say. Here is a fairlycomplete answer. Since X ∈ {X,∅} Property 1 is satisfied.

6 JOHN HILLAS UNIVERSITY OF AUCKLANDSince ∅ ∈ {X,∅} Property 2 is satisfied. For any A,B ∈{X,∅} if A = B = X then A∩B = X ∈ {X,∅}. Otherwiseat least one of A and B are ∅ and then A∩B = ∅ ∈ {X,∅}.So, in all cases, if A,B ∈ {X,∅} then A ∩ B ∈ {X,∅} soProperty 3 is satisfied. Now suppose that for some collectionof sets {A n | n ∈ N} with A n ∈ {X,∅} for all n. Each A n iseither ∅ or X. If A n = ∅ for all n then ∪ n∈N A n = ∅ ∈ {X,∅}.If not then A n ≠ ∅ for at least one n and for that n it must bethat A n = X. But then ∪ n∈N A n = X ∈ {X,∅}. So Property4 is satisfied.(iii) Show that this topology cannot come from any metric. (Hint: you haveessentially already show this in your answer to part (i).)In part (i) we saw that whatever metric we use the collectionof open sets is all subsets of {a,b,c,d,e}. There are 32 suchopen sets. The topology τ = {X,∅} is not this collection soit cannot come from any metric.(iv) Give another set τ that satisfies the four properties given above.There are many such τ. Here are a few examples:τ = {X,{a},∅}τ = {X,{a,b},∅}τ = {X,{a},{a,b},∅}τ = {X,{a},{b},{a,b},∅}τ = {X,{a},{b},{a,b},{a,b,c},∅}τ = {X,{c},{a,b,c},{c,d,e},∅}(v) Characterise the set of all possible topologies τ on X that satisfy the fourproperties.Looking at the examples I gave in the previous part you seesome difficulties if you try to give a very explicit characterisationvery far away from simply “applying the properties.”Here is one characterisation. There are 32 subsets of X.The collection must include ∅ and X. You can include anyof the other 30 subsets. Then add all intersection of subcollectionsof the sets you have added and all unions of subcollectionsof the sets you have added. The resulting collectionwill satisfy the axioms and all collections satisfying theaxioms can be obtained in this way.